# Chapter 4 - Motion in Two and Three Dimensions - Problems: 16b

t = 0.75s

#### Work Step by Step

In the previous section we found that a = $(6-8t)i$ Therefore, to find when the acceleration is 0, we plug in 0 for a and solve for the t at which this occurs. The j and k components are always 0 so we need only worry about the i component. The magnitude of this vector is given by: $magnitude = \sqrt(i^2 + j^2 +k^2)$ where i, j, and k represent the i, j, and k components respectively Therefore, we have $magnitude = \sqrt((6-8t)^2)$ which is just 6-8t Therefore we can solve for when a = 0: $0 = 6-8t$ so then t = 0.75s

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